# Proving that an integral is differentiable

Given an open interval $I\subset\mathbb{R}$, $i_0\in I$, and $f:I\times\mathbb{R} \rightarrow\mathbb{R}$ such that

1. For all $i\in\mathbb{R}:y\mapsto f(i,y)$ is integrable

2. For all $y \in \mathbb{R}$ the partial derivative $\frac{\partial f}{\partial i}(i_0,y)$ exists and

3. T.e. a neighbourhood $V$ of $i_0$ and an integrable function $h:\mathbb{R}\rightarrow\mathbb{R}$ s.t. for all $i \in V\cap I, i\neq i_0$ and for all $y\in\mathbb{R}$ we have $\frac{f(i,y)-f(i_0,y)}{i-i_0}\le h(y)$

I want to prove that $$F:I\rightarrow\mathbb{R}, i\mapsto\int_{\mathbb{R}}f(i,y)dy$$ is differentiable at $i_0$ and that $$F'(i_0)=\int_{\mathbb{R}}\frac{\partial f}{\partial i}(i_0,y)dy$$

I know that for $F$ to be differentiable I have to prove that $$\lim_{i\rightarrow i_0}\frac{F(i)-F(i_0)}{i-i_0}$$ exists. If this turns out to be the expression from above, I'm done I think. So at the outset I have $$\frac{F(i)-F(i_0)}{i-i_0}=\frac{\int_{\mathbb{R}}f(i,y)dy-\int_{\mathbb{R}}f(i_0,y)dy}{i-i_0}=\frac{\int_{\mathbb{R}}(f(i,y)-f(i_0,y))dy}{i-i_0}$$

What do I do next?

• Essentially, you want to show that the difference quotients $$\frac{F(i+h) - F(i)}{h} = \int_{X} \frac{f(y, i+h) - f(y, h)}{h} dy$$ converge as $h \to 0$, which means you need to be able to pass to the limit inside the integral. Do you know how you can show that? – Mattos Apr 12 '16 at 14:40
• Do you mean $f(y,i)$ instead of $f(y,h)$ on the right side of the equation? And no, sadly i don't. – azureai Apr 12 '16 at 16:01
• Yes, I meant $f(y, i)$ in the integral. I'll write a proof when I get home. Alternatively, you are looking for a proof of differentiation of parameter integrals, which you may be able to find online. – Mattos Apr 13 '16 at 3:00
• I didn't really find anything of use, it would be nice if you could write something later. – azureai Apr 13 '16 at 16:15

Fix an $i \in I$ and define the function

$$u(y,h) := \begin{cases} \frac{f(y,i+h) - f(y,i)}{h}, & \text{if } h \ne 0 \\\\ \frac{\partial f}{\partial i}(y,i), & \text{if } h = 0 \end{cases}$$

From our conditions, this function is well defined for almost all $y \in \mathbb{R}$ and since the partial derivative exists, $i \mapsto u(y,i)$ is continuous on $I$ for almost all $y \in I$. Also, $i \mapsto f(y,i)$ is continuously differentiable, so the mean value theorem implies the existence of a $c \in I$ with $\lvert c - i \rvert < \lvert h \rvert$ such that

\begin{align} u(x,h) &= \frac{f(y,i+h) - f(y,i)}{h} \\ &= \frac{\partial f}{\partial c} (y,c) \end{align}

Hence, by assumption

\begin{align} \lvert u(x,h) \rvert &= \bigg \lvert \frac{\partial f}{\partial c} (y,c) \bigg \rvert \\ &\le g(y) \end{align}

(I used $g(y)$ not $h(y)$ to differentiate the function from the step size $h$) for almost all $y \in \mathbb{R}$ and all $h \in \mathbb{R}$ with $i + h \in I$. So the requisite assumptions are satisfied and so

\begin{align} F'(i) &= \lim_{h \to 0} \frac{F(i+h) - F(i)}{h} \\ &= \lim_{h \to 0} \int_{\mathbb{R}} u(x,h) dy \\ &=\int_{\mathbb{R}} u(y,0) dy \\ &= \int_{\mathbb{R}} \frac{\partial f}{\partial i} (y,i) dy \end{align}

• I can't quite follow. What happened to the $i_0$ from the opening post? – azureai Apr 13 '16 at 17:34
• "Fix an $i \in I$".. That is your $i_{0}$. – Mattos Apr 14 '16 at 1:54
• Thank you for your answer. But do I really have to work with 'almost everywhere' etc.? Because I still don't understand your proof to be honest. I could probably work out the details on my own but what is the main idea in your proof? How does this have to do with what you wrote in your first comment on the opening post, that is: 'the difference quotients converging'? – azureai Apr 14 '16 at 17:51
• And what is $g$? – azureai Apr 14 '16 at 18:08

Fix an $i \in \mathbb{R}$ and define the function

$$u(y,h) := \begin{cases} \frac{f(y,i+h) - f(y,i)}{h}, & \text{if } h > \ne 0 \\\\ \frac{\partial f}{\partial i}(y,i), & \text{if } h = 0 > \end{cases}$$

From our conditions, this function is well defined for almost all $y > \in \mathbb{R}$ and since the partial derivative exists, $i \mapsto > u(y,i)$ is continuous on $I$ for almost all $y \in I$. Also, $i > \mapsto f(y,i)$ is continuously differentiable, so the mean value theorem implies the existence of a $c \in I$ with $\lvert c - i \rvert > < \lvert h \rvert$ such that

-> the opening post was f(i,y), does it mean $y \in R$, doesn't it? and the variable h is equivalent to which one I saw in your post they are all f(y,i), but in the opening post was f(i,y)

\begin{align} u(x,h) &= \frac{f(y,i+h) - f(y,i)}{h} \\ &= \frac{\partial f}{\partial c} (y,c) \end{align}

Hence, by assumption

\begin{align} \lvert u(x,h) \rvert &= \bigg \lvert \frac{\partial f}{\partial c} (y,c) \bigg \rvert \\ &\le g(y) \end{align}

(I used $g(y)$ not $h(y)$ to differentiate the function from the step size $h$) for almost all $y \in \mathbb{R}$ and all $h \in \mathbb{R}$ with $i + h \in I$. So the requisite assumptions are satisfied and so

\begin{align} F'(i) &= \lim_{h \to 0} \frac{F(i+h) - F(i)}{h} \\ &= \lim_{h \to 0} \int_{\mathbb{R}} u(x,h) dy \\ &=\int_{\mathbb{R}} u(y,0) dy \\ &= \int_{\mathbb{R}} \frac{\partial f}{\partial i} (y,i) dy \end{align}

and what happens to i_0?